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Suppose $\{e_1,...,e_N\}$ is the set of all extreme points of a compact convex subset $X\subset\mathbb R^n$. $L: \mathbb R^n\to \mathbb R^m$ is a linear transformation. $L$ is surjective but is not injective. Let $Y= L(X)$.

Would it hold that for every $1\leq i\leq N$, $L(e_i)$ must be an extreme point of $Y$? Is there any characterization on $L$ such that this property holds?

Thanks.

user91360
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2 Answers2

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If $L$ fails to be injective, then this will not hold.

For example, take $X \subset \Bbb R^2$ to be $\{(x,y):|x|+|y| \leq 1\}$, and take $L:\mathbb{R}^2 \to \mathbb{R}$ to be given by $L(x,y) = x$. Note that $(0,1)$ is an extreme point of $X$, but $L(0,1) = 0$ is not an extreme point of $Y$.

On the other hand, if $L$ is injective in addition to being surjective, then $L$ is an invertible linear transformation and so the statement holds.

RB1995
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Ben Grossmann
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  • FIRST I apologise as I am using this for the first time and it would not let me comment, only edit. But I did not want to edit what was written above and so nothing was changed in the above post and I just add something here. I had a follow up question. You wrote if L is not inj, then this (meaning extremal points will be mapped to extremal points) will not hold and gave an example. That only shows that L if noninj. then the above condition MIGHT fail. Is it always the case that noninj. – Chill2Macht Jan 17 '17 at 13:18
  • implies that some extremal guy is sent to a nonextremal guy? Is this easy? Reference? True in infinite dimensions? Thanks...Ping – Chill2Macht Jan 17 '17 at 13:18
  • @William It's not always the case. For the above $L$, take any rectangle with sides parallel to the axes. – Ben Grossmann Jan 17 '17 at 13:55
  • @William you probably would have gotten a quicker answer if you asked a question( in which you linked to my post). – Ben Grossmann Jan 17 '17 at 13:56
  • The new user "Ping" tried to edit your post since they didn't have enough reputation to comment, so I rejected the edit and copy/pasted their comment here. Probably not the optimal solution, but it didn't seem like a spam comment so it did not seem fair to reject the edit and not save the attempted comment. – Chill2Macht Jan 17 '17 at 14:31
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    @William I misunderstood! I thought you were telling me to "ping" you, which is to say type "@William" in the front. That makes sense now. – Ben Grossmann Jan 17 '17 at 14:52
  • @Ping see the comments. Also, you should have posted a question. – Ben Grossmann Jan 17 '17 at 14:53
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The images by $L$ of the extreme points are certainly not all extreme points of the image of the compact convex by $L$.

Take the example of a square and for $L$ the projection on one of the diagonal. The two extreme points on the diagonal are also extreme points of the image by $L$ of the square, but that is not the case for the opther two vertex of the square.

However the extreme points of the image of the compact convex is the subset of the images of the extreme points of the initial compact convex.

mathcounterexamples.net
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